In Delta PQR,m angle Q = 90^{circ} and overli | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) In a right-angled triangle $\Delta PQR$ with $\angle Q = 90^{\circ}$,$\overline{QM}$ is the altitude to the hypotenuse $\overline{PR}$.By the geom

Vedclass · Accepted Answer

$2$ or $8$

Vedclass · Answer

(D) In a right-angled triangle $\Delta PQR$ with $\angle Q = 90^{\circ}$,$\overline{QM}$ is the altitude to the hypotenuse $\overline{PR}$.By the geometric mean theorem (or properties of similar triangles),we have $QM^2 = PM \cdot MR$.Let $PM = x$. Since $PR = 10$,then $MR = 10 - x$.Given $QM = 4$,we substitute these values into the equation:$4^2 = x(10 - x)$$16 = 10x - x^2$$x^2 - 10x + 16 = 0$Factoring the quadratic equation:$(x - 2)(x - 8) = 0$Thus,$x = 2$ or $x = 8$.Therefore,$PM$ can be $2$ or $8$.

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QM}$ is an altitude to the hypotenuse $\overline{PR}$. If $QM = 4$ and $PR = 10$,find $PM$.

Similar Questions

In the figure,altitudes $AD$ and $CE$ of $\Delta ABC$ intersect each other at the point $P$. Show that $\Delta PDC \sim \Delta BEC$.

$\Delta ABC \sim \Delta XYZ$ for the correspondence $ABC \leftrightarrow ZYX$. If $AB = 3, BC = 5, CA = 6$ and $XY = 6$,find the perimeter of $\Delta XYZ$. (in $.8$)

In $\Delta GBS$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{GS}$. Prove that $\frac{GB^{2}}{BS^{2}} = \frac{GM}{SM}$.

If in two triangles $DEF$ and $PQR$,$\angle D = \angle Q$ and $\angle R = \angle E$,then which of the following is not true?

Given $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. If $2 m \angle P = 3 m \angle Q$ and $m \angle C = 100^{\circ}$,find $m \angle B$. (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module